91Ó°ÊÓ

We should find oriented with its positive terminal at the top or at the bottom.

When the current flows through a resistor, it dissipates energy, the rater of the dissipated energy is the power. This rate where the energy is transferred to the current to the resistor is

$P_R=I∆V_R$

The dissipated power is zero when the current through the resistor $200\mathrm{Î\copyright }$is zero (no current flow). But at the condition, the current will not flow through the $200\mathrm{Î\copyright }$? This occurs when the right battery $\mathrm{Î\mu }$applies a current through $300\mathrm{Î\copyright }$with the same magnitude of the left battery but in opposite direction. So, let us find the current that is flow due the left battery $50V$using Ohm's law.

Ohm's law states that the current Iflows through a resistance due to the potential difference V between resistance.

$I=\frac{V}{R}$

So, we get the current through $100\mathrm{Î\copyright }$by using the values $50Vand100\mathrm{Î\copyright }$into equation $\left(2\right)$

$I_{100\mathrm{Î\copyright }}=\frac{V}{R}=\frac{50V}{100\mathrm{Î\copyright }}=0.5A$

As we mentioned above, this is the current that flows in $300\mathrm{Î\copyright }$to make no current flows through $200\mathrm{Î\copyright }$. So, the voltage that exert this current through the resistor $300\mathrm{Î\copyright }$is calculated by Ohm's law in the form

$$\begin{gathered} \mathrm{Î\mu }=IR \\ =(0.5A)\left(300\mathrm{Î\copyright }\right) \\ =150V \end{gathered}$$

The right battery has emf $\mathrm{Î\mu }=150V$and to exerts a current in the opposite direction, its terminal positive should be bottom.